high-level language structures

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High-level language structures

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Branching structuresIF-THEN

• IF condition is true

THEN

execute true-branch statements

End_if

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Branching structuresIF-THEN

false truecondition

True branchstatements

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example

• Replace the number in AX by its absolute value.

• Pseudo code algorithm is

IF AX <0

Then

replace AX by –AX

END_IF

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example

• Code:

CMP AX,0 ;AX<0?

JNL END_IF ;no, exit

NEG AX ;yes, change sign

END_IF:

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IF_THEN_ELSE

• IF condition is true

THEN

execute true-branch statements

ELSE

execute FALSE-branch statements

End_IF

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IF_THEN_ELSE

FALSE TRUECONDITION

TRUE BRANCHSTATEMENTS

FALSE BRANCHSTATEMENTS

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CASE• A case is a multiway branch structure that

tests a register, variable, or expression for particular values or range of values. The general form is as follows:

case expression values_1: statements_1 values_2: statements_2 ………….. values_n: statements_n

END_CASE

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CASE

EXPRESSION

STATEMENTS_1 STATEMENTS_2 STATEMENTS_N

VALUE_1 VALUE_2 VALUE_N…….

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EXAMPLE

• IF AX contains a negative number, put -1 in BX; if AX contains 0, put 0 in BX; if AX contains a positive number, put 1 in BX.

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example

• Pseudo code algorithm is

Case AX

< 0:put -1 in BX

= 0:put 0 in BX

> 0:put 1 in BX

END_CASE

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ExampleCode: CMP AX,0 JL NEGATIVE JE ZERO JG POSITIVE NEGATIVE: MOV BX,-1 JMP END_CASE

ZERO: MOV BX,0 JMP END_CASEPOSITIVE: MOV BX,1END_CASE

•NOTE: only one CMP is needed, because jump instructions do not affect the flags.

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example

• If AL contains 1 or 3, display “o”; if AL contains 2 or 4, display “e”.

• Pseudo code algorithm is

• Case AL

1, 3: display “o”

2, 4: display “e”

End_case

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example

Code:CMP AL,1

JE ODD

CMP AL,3

JE ODD

CMP AL,2

JE EVEN

CMP AL,4

JE EVEN

JMP END_

ODD:

MOV DL, ’o’

JMP DIS_

EVEN:

MOV DL, ’e’

END_:

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Branches with compound conditions

• Sometimes the branching condition in an IF or CASE takes the form

• Condition_1 and condition_2

• Condition_1 or condition_2

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AND conditions

• An AND condition is true if and only if condition_1 and condition_2 are both true.

• Likewise, if either condition is false, then the whole thing is false.

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Example

• Read a character, and if it’s an uppercase letter, display it.

• Pseudo code algorithm is

read a character (into AL)

If (‘A’ <= character) and (character <=‘Z’)

Then

Display character

End_if

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example

• Code:

MOV ah,1

INT 21h

CMP al,’A’

JL end_if

CMP al,’Z’

JG end_if

MOV DL,AL

MOV AH,2

INT 21H

End_if:

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OR conditions

• condition_1 OR condition_2 is true if at least one of the conditions is true.

• if both conditions are false, then the whole thing is false.

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example

• Read a character. If it’s “y” or “Y”, display it; otherwise, terminate the program.

• Pseudo code algorithm is Read a character (into AL) if (character = “y” or character= “Y”) Then Display character Else Terminate the program End_if

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example

MOV AH,1 INT 21h CMP AL, ’y’ JE THEN CMP AL,’Y’ JE THEN JMP ELSE_

THEN: MOV AH,2 MOV DL,AL INT 21H JMP END_IF ELSE_:MOV AH,4CHINT 21HEND_IF:

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Looping structures

• A loop is a sequence of instructions that are repeated.

• For loop:

this is a loop structure in which the loop statements are repeated a known number of times (a count-controlled loop).

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For loop_count times do statementsend_for

Initialize count

statements

Count = count - 1

Count=0False

True

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LOOP instruction

• The LOOP instruction can be used to implement a FOR loop.

• General form:

LOOP des_label

The counter for the loop is the register CX.

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LOOP instruction• CX is initialized to loop_count.• Execution of the LOOP instruction causes

CX to be decremented automatically.• If CX is not 0, control transfers to

des_label.• If CX =0, the next instruction after LOOP is

done.• Des_label must precede the LOOP

instruction by no more than 126 bytes.

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LOOP instruction

• So, a FOR loop can be implemented as follows.

; initialize CX to loop_count

Des_label:

; body of the loop

loop des_label

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example

• Write a count-controlled loop to display a row of 80 stars.

• Pseudo code algorithm is

For 80 times do

display ‘*’

end_for

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example

MOV CX,80

MOV AH,2

MOV DL,’*’

TOP:

INT 21H

LOOP TOP

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LOOP instruction

• A loop instruction is executed at least once.

• If CX = 0 when the loop is entered, the loop instruction causes CX to be decremented to FFFFH.

• So, the loop is executed 65535 more times.

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LOOP instruction

• Solution:

use the instruction JCXZ (jump if CX is zero) before the loop.

JCXZ des_label

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LOOP instruction

• A loop implemented as follows is bypassed if CX is 0.

JCXZ skip

top:

;body of the loop

loop top

Skip:

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While Loop

• This loop depends on a condition.

• Pseudo code

while condition do

statements

end_while

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While Loop

statements

conditionTrue

False

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example• Write some code to count the number of

characters in an input line.

• Pseudo code

initialize count to 0.

read a character

while character <> carriage_return do

count=count + 1

read a character

end_while

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example MOV DX,0 MOV AH,1 INT 21HWHILE_: CMP AL,0DH JE END_ INC DX INT 21H JMP WHILE_END_ :

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Repeat loop

• Another conditional loop is the repeat loop. Pseudo code

repeat

statements

until condition

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statements

condition

True

False

Repeat loop

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example

• Write some code to read characters until a blank is read.

Pseudo code

Repeat

read a character

until character is a blank

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example

MOV AH,1

REPEAT:

INT 21H

CMP AL,’ ’

JNE REPEAT

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WHILE versus REPEAT

• In a while loop, the loop can be bypassed if the terminating condition is initially false.

• The statements in a repeat must be done at least once.

• Code for a repeat loop is likely to be shorter because there is only a conditional jump at the end, but a while loop has two jumps: a conditional jump at the top and a jmp at the bottom.

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Programming with high-level structures

• Top-down program design:

The original problem is solved by solving a series of sub problems, each of which is easier to solve than the original problem.

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examplePrompt the user to enter a line of text.On the next line, display the capital letter

entered that comes first alphabetically and the one that comes last. If no capital letters are entered, display “No capital letter”.

sample execution: type a line of text: THE QUICK BROWN FOX JUMPED. first capital = B last capital = X

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Top down program design

First refinement:

1. Display the opening message.

2. Read and process a line of text.

3. Display the results.

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example TITLE PGM6_2 : FIRST LAST CAPITALS.MODEL SMALL.STACK 100H.DATAPROMPT DB 'Type a line of text ',0dH,0AH,'$'NOCAP_MSG DB 0DH,0AH,' No Capitals $'CAP_MSG DB 0DH,0AH,'FIRST CAPITAL = 'FIRST DB '[' DB ' LAST CAPITAL = 'LAST DB '@ $'

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.CODEMAIN PROC; initialize DS MOV AX,@DATA MOV DS, AX; first step display opening message MOV AH,9 LEA DX,PROMPT INT 21H

Step 1

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Step 2: read and process a line of text Read a character.While character is not a carriage return doIf character is a capital letter Then If character precedes first capital Then First capital = character End_if If character follows last capital Then Last capital = character End_ifEnd_ifRead a characterEnd_while

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;read and process a line of text MOV AH,1 INT 21HWHILE_:; while character is not carriage ;return do CMP AL,0dH JE END_WHILE; if character is a capital letter CMP AL,'A' JNGE END_IF CMP AL,'Z' JNLE END_IF; then; if character precedes FIRST ;capital CMP AL,FIRST JNL CHECK_LAST

; then FIRST CAPITAL = ;character MOV FIRST,AL; end_ifCHECK_LAST :; if character follows last capital CMP AL,LAST JNG END_IF; then LAST CAPITAL = character MOV LAST,AL;end_ifEND_IF:; read a character INT 21H JMP WHILE_END_WHILE :

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Step 3: display the results

if no capitals were typed

then

display “no capitals”

Else

display first capital and last capital

End_if

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; display results MOV AH,9; if no capital were typed CMP FIRST, '[' JNE CAPS; then LEA DX,NOCAP_MSG JMP DISPLAYCAPS : LEA DX,CAP_MSGDISPLAY : INT 21H;end_if; DOS exit MOV AH,4CH INT 21HMAIN ENDP END MAIN